A triangle has corners A, B, and C located at #(4 ,2 )#, #(2 ,6 )#, and #(8 ,4 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Question

A triangle has corners A, B, and C located at #(4  ,2  )#, #(2  ,6    )#, and #(8 ,4   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Aurora Ayala · Accepted Answer

Length of altitude #=2\sqrt5# & end-points of altitude #(8, 4)# & #(4, 2)#

The vertices of #\Delta ABC# are #A(4, 2)#, #B(2, 6)# & #C(8, 4)#

The area #\Delta# of #\Delta ABC# is given by following formula

#\Delta=1/2|4(6-4)+2(4-2)+8(2-6)|#

#=10#

Now, the length of side #AB# is given as

#AB=\sqrt{(4-2)^2+(2-6)^2}=2\sqrt5#

If #CN# is the altitude drawn from vertex #C# to the side #AB# then the area of #\Delta ABC# is given as

#\Delta =1/2(CN)(AB)#

#10=1/2(CN)(2\sqrt5)#

#CN=2\sqrt5#

Let #N(a, b)# be the foot of altitude CN drawn from vertex #C(8, 4)# to the side #AB# then side #AB# & altitude #CN# will be normal to each other i.e. the product of slopes of AB & CN must be #-1# as follows

#\frac{b-4}{a-8}\times \frac{6-2}{2-4}=-1#

#a=2b\ ............(1)#

Now, the length of altitude CN is given by distance formula

#\sqrt{(a-8)^2+(b-4)^2}=2\sqrt5#

#(2b-8)^2+(b-4)^2=(2\sqrt5)^2#

#(b-4)^2=4#

#b=6, 2#

Setting these values of #b# in (1), we get the corresponding values of #a# as follows

#a=2\times 6=12\ \ & \ \ \ a=2(2)=4#

#a=12, 4#

The endpoints of altitude from vertex #C(8, 4)# are #(12, 6)# & #(4, 2)# But #(12, 6)# is not the end point of altitude.

hence, the end points of altitude from vertex #C# are

#(8, 4), (4, 2)#

Aaliyah Anderson · Answer

#A(4,2) and C(8,4) # are the endpoints of altitude.

Length of the altitude #=AC=2sqrt5#

Let , #triangle ABC" be the triangle with corners at"#

#A(4,2) , B(2,6) and C(8,4)#

#"Using "color(blue)"Distance formula :"#

Distance between two points #P(x_1,y_1) and Q(x_2,y_2)# is :

#color(blue)(PQ=sqrt((x_1-x_2)^2+(y_1-y_2)^2)#

#AB=sqrt((4-2)^2+(2-6)^2)=sqrt(4+16)=sqrt20#

#BC=sqrt((8-2)^2+(4-6)^2)=sqrt(36+4)=sqrt40#

#AC=sqrt((4-8)^2+(2-4)^2)=sqrt(16+4)=sqrt20#

#i.e. (AB)^2=20 , (BC)^2=40 and (AC)^2=20#

#:.(AB)^2+(AC)^2=(BC)^2=40#

#:.triangle ABC " is Right Triangle" =>mangle A=90^circ#

So , #A(4,2) "is the Orthocenter of" triangle ABC"#

It is clear that , #bar(AC)# is the altitudes of #triangle ABC ,# going through

corner #C(8,4).#

So, #A(4,2) and C(8,4) # are the endpoints of altitude.

Length of the altitude #=AC=sqrt20=2sqrt5#

James Atkins · Answer

undefined The endpoints of the altitude going through corner C are the coordinates of C and the foot of the altitude, which is the point where the altitude intersects the side opposite corner C. The length of the altitude can be found using the distance formula between the endpoints.

To find the foot of the altitude, we need to find the equation of the line passing through point C that is perpendicular to the line containing the segment AB. Then, we find the intersection point of this perpendicular line with segment AB.

Let's denote the coordinates of point A as (x1, y1), point B as (x2, y2), and point C as (x3, y3).

Given:
A (4, 2)
B (2, 6)
C (8, 4)

The equation of the line containing AB:
$$ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) $$

The slope of AB:
$$ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} $$

Perpendicular slope:
$$ m_{\perp} = -\frac{1}{m_{AB}} $$

Using point-slope form, the equation of the line perpendicular to AB passing through C:
$$ y - 4 = -\frac{1}{3}(x - 8) $$

Now, we solve this equation simultaneously with the equation of line AB to find the foot of the altitude.

After finding the foot of the altitude, we can calculate the length of the altitude using the distance formula between the endpoints (C and the foot of the altitude).